Ether channel

[jcrosby.vol11]

When using the show interface port-channel # etherchannel command. What does the load categorie represent?

[Georg Pauwen]

Hello,

the 'load' is the RBH (Result Bundle Hash) value, which basically tells you which packets use which port in the channel.

The concept looks a bit confusing at first, but is actually pretty simple:

If you have a channel made up of two ports, the 'load' values are typically 'AA' and '55'. These are dual HEX numbers. If you convert these to binary, you will get the following:

A = 1010

A = 1010

So, AA = 10101010

5 = 0101

5 = 0101

And 55 = 01010101

The RBH values are 0 through 7. 

76543210

If you match the binary numbers against these values, 10101010 (or HEX AA) means that the values for 1,3,5, and 7 are set

76543210

10101010

For the binary number 01010101 (or HEX 55), the values 0,2,4, and 6 are set:

76543210

01010101

That means that with two ports in the channel, the load is equally balanced (50:50). With three or four ports in the channel, you will get a different distribution.

Does this make sense ? I hope it does...

[Jose Manuel Garcia]

It's pretty hard to understand. At least, for me.

I'm taking CCNA exam next month and I have a great doubt about this item. Specifically, I found this question that I can't answer. Can you help me?

While troubleshooting a switch, you executed the “show interface port-channel 1 etherchannel” command and it returned this output. Which information is provided by the Load value?

A. the percentage of use of the link
B. the preference of the link
C. the session count of the link
D. the number source-destination pairs on the link

What do you think is the right answer and why?

Thank you in advanced

[Mark Malone]

This is maybe be easier to understand regarding the load and how it works well explained doc

http://packetpushers.net/understand-etherchannel-load-balancing-catalyst-switches/

[Jose Manuel Garcia]

Thank you for the link.

It is not very clear yet, but if I don't misunderstand, the correct answer would be D...

Anyway, I'll reread the article once again.